Anonymous user
Combinations with repetitions: Difference between revisions
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Line 2,443:
: (length (combrep 3 (range 1 10)))
-> 220</pre>
=={{header|Prolog}}==
<lang prolog>
combinations_of_length(_,[]).
combinations_of_length([X|T],[X|Combinations]):-
combinations_of_length([X|T],Combinations).
combinations_of_length([_|T],[X|Combinations]):-
combinations_of_length(T,[X|Combinations]).
</lang>
<pre>
?- [user].
|: combinations_of_length(_,[]).
|: combinations_of_length([X|T],[X|Combinations]):-
|: combinations_of_length([X|T],Combinations).
|: combinations_of_length([_|T],[X|Combinations]):-
|: combinations_of_length(T,[X|Combinations]).
|: ^D% user://1 compiled 0.01 sec, 3 clauses
true.
?- combinations_of_length([0,1,2,4],[L0, L1, L2, L3, L4, L5]).
L0 = L1, L1 = L2, L2 = L3, L3 = L4, L4 = L5, L5 = 0 ;
L0 = L1, L1 = L2, L2 = L3, L3 = L4, L4 = 0,
L5 = 1 ;
L0 = L1, L1 = L2, L2 = L3, L3 = L4, L4 = 0,
L5 = 2 ;
L0 = L1, L1 = L2, L2 = L3, L3 = L4, L4 = 0,
L5 = 4 ;
L0 = L1, L1 = L2, L2 = L3, L3 = 0,
L4 = L5, L5 = 1 ;
L0 = L1, L1 = L2, L2 = L3, L3 = 0,
L4 = 1,
L5 = 2 ;
L0 = L1, L1 = L2, L2 = L3, L3 = 0,
L4 = 1,
L5 = 4 ;
L0 = L1, L1 = L2, L2 = L3, L3 = 0,
L4 = L5, L5 = 2 ;
L0 = L1, L1 = L2, L2 = L3, L3 = 0,
L4 = 2,
.
.
.
</pre>
=={{header|PureBasic}}==
Line 2,526 ⟶ 2,569:
Ways to select 3 items from 10 types: 220</pre>
=={{header|Python}}==
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